Diamond Solitaire

Diamond Soli taire George I. Bell Septem b er 2005 1 gibell@comc ast.net Abstract W e in v estigate the game of peg solita ire on differen t bo ard shap es, and find those of diamond or rhom bus shap e ha v e in teresting prop erties. When one p eg captures man y p egs consecu- tiv ely , this is called a swe ep. Rhom bus b oards of side 6 hav e the prop ert y that no matter whic h p eg is missing at the start, the game can b e solv ed to one p eg using a maximal sw eep of length 16. W e sho w how to construct a solution on a rhom bus b oard of side 6 i , where the final mov e is a maximal sw eep of length r , where r = (9 i − 1)(3 i − 1) is a “rhom bic matc hstic k nu mber”. 1 In tro ducti on P eg solitaire is a o ne- p erson game usually play ed on a 33- ho le cross-shap ed b oard, or a 15 - hole triangular b oard (Figure 1). In the first case the pattern of holes (or b oa rd lo cations) come from a square lat t ice, while in the second case they come from a triangular lattice. The usual game b egins with the b oard filled b y p egs (or marbles) except for a single b oard lo cation, left v acan t. The pla y er then jumps one p eg o ve r another in to an em pty h o le, remo ving the jump ed p eg from the b oard. The goal is to c ho ose a series of jumps that finish with one p eg. The general problem of going from a b oard p o sition with one p eg missing to one p eg will b e called a p eg solitaire problem . If the missing p eg and final p eg are in the same place, w e call it a complemen t problem , b ecause the starting and ending b oard p ositions are complemen ts o f one another (where ev ery p eg is replaced b y an empt y hole and vice v ersa). Initially , w e will consider p eg solitaire on b oa rds of rather arbitrary shap e. The basic problem ma y seem hard enough, but w e will a dd more conditions or constrain ts on the solution. The reason for this may not b e immediately clear, but a dding constrain ts to a pro blem with man y solutions can mak e it easier to find one, a nd the solutions themselv es can b e quite remark able. The approac h also brings o ut certain b oa rd shap es with in teresting prop erties. 1 Original version at ht tp:// gpj.co nnectfree.co.uk/gpjw.htm Conv erted to L A T E X by the author with some mo dificatio ns to the text, No vem b er 2007. The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 2 Supp ose w e add the constrain t that the p eg solitaire problem m ust finish in the most dramatic w a y p ossible: with one p eg “swe eping off” all the remaining p egs, and finishing a s the sole surviv or. Here we need to distinguish b et w een a jump —one p eg jumping o v er another, and a mov e —one o r more jumps by the same p eg. An y single mo ve that captures n p egs is a sw eep of length n , or an n -sw eep . In this terminolog y w e w an t to finish a p eg solitaire problem with the longest sw eep p ossible. A sw eep that has the long est length geometrically p ossible on a b oard is called a maximal sw eep. F or most b oard shap es, maximal sw eeps cannot b e reac hed in the solution to a p eg solitaire problem. In GPJ #36 [6], w e considered the triangular b oa r d of side n , called T riangle ( n ). W e sa w that for n o dd, the boa rd T r ia ngle ( n ) supp ort s an even more remark able sw eep that is maximal and also has the prop erty t ha t ev ery hole in the b oard is affected b y the sw eep in that it is either jump ed o v er or is the starting or ending ho le of some jump in the sw eep. Suc h remark able sw eeps need a sp ecial na me—w e call them sup er-swee ps . (a) (b) Figure 1 : Maximal swe eps on the standard b o ards, only the second is a sup er- swe ep. The sp ecial green (or shaded) p eg p erforms the sw eep. (a ) A 16-sw eep on the 33-hole cross-shap ed b oard. (b) A 9-sw eep on the 15- hole triangular b oa rd, T riangle (5). In Figure 1a, the cen tral hole is not to uched or jump ed o ve r by the sw eep, thus while this sw eep is the longest p o ssible it is not a sup er- sw eep, and a sup er- swe ep is not p ossible on this b oa rd. It’s not hard to see that sup er-swe eps are nev er p ossible o n square lattice b oards, except f o r certain trivial or degene ra te cases (like a 1-dimensional b oard). Non-trivial sup er- sw eeps a r e only p ossible on a triangular lattice, an example is Figure 1 b. Neither of the sw eep patterns of Figure 1 can b e reac hed during the solution to a p eg solitaire problem o n that b oar d. The easiest w ay to see this is to try to figure out ho w one could ha v e arriv ed at t he sw eep b oard p osition, or equiv alen tly try to play b ackwar ds from the sw eep p osition. In GPJ #36 [6], we sa w in the F orw ard/Bac kw ard Theorem that bac kw ard pla y is equiv alent to fo rw ard pla y from the complemen t of the b oard p osition. This concept is referred to as the “time rev ersal trick ” in Winning W a ys F or Y our Mathematical Pla ys [1]. If we tak e the complemen t of either b oard p osition in Figure 1, w e find that no jump is possible. This prov es tha t there is no “paren t b oard position” from whic h the s we ep p osition can a rise, it canno t occur during the solution to a p eg s olita ire problem. If w e tak e t he complemen t o f an y sup er-sw eep pattern, and a jump is p ossible, this means there w ould ha ve to b e t w o consecutiv e empt y holes in the original super- sw eep. Because it is a sup er-sw eep, b oth of these holes m ust b e the starting or ending lo cations of some jump in The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 3 the sw eep, whic h is imp ossible. Therefore a sup er-sw eep c an never o c cur in the solution to a p eg solita ire problem. If sup er-sw eeps cannot o ccur in p eg solitair e problems, the reader ma y w onder why we are w asting our time with them. The a nsw er is provided b y the tria ngular b oards and G PS #36 [6]: the sup er-swe ep pattern of Figure 1b c an b e reach ed in a problem on T riangle (6). This 9 -sw eep is no longer a sup er-sw eep with r esp ect to T riangle (6), b ut it is s til l a maxim al swe ep . Lo osely sp eaking, a sup er-sw eep may still b e reac hable in a p eg solitaire problem on a b oard one size larger 2 . 2 General b oards with sup er- sw eeps Let us consider p eg solitaire b oards on a tria ngular lattice where the bo ard shap e is a general p olygon. On a triangular la ttice the cor ners of the b oa r d are restricted to m ultiples of 60 ◦ . F or whic h suc h b oards is it p ossible to hav e a sup er-swe ep? It is clear that the b oard edges m ust all hav e an o dd length (in holes), b ecause the sup er-sw eep m ust pass thro ugh all the corners of the b oard (b y definition, a corner hole cannot b e jump ed o ve r, a nd consequen tly the super- sw eep mu st jump into it). Consider, fo r example, the 8-sided po lygon bo ard of Figure 2. (a) (b) Figure 2: An 8- sided (non- con v ex) p olygon b oa r d and the asso ciated sup er-swe ep g raph. (a) An un usual 24-ho le b oard, set up in the b oard p osition of a h yp othetical sup er-swe ep ( of length 15). (b) The graph of the h yp othetical sup er-sw eep. Can the green p eg (top left) in F igure 2a mak e a t o ur of the b oard, sw eeping off all 15 remaining p egs? W e can answ er this question b y lo oking at the graph f ormed by the sw eep, sho wn separately in Figure 2b rather than on top of the b oard as in Figure 1. In the language of graph theory the sup er-swe ep is an Euler path on this graph, i.e. a pat h that trav ers es ev ery edge exactly once. One of the most basic theorems of graph theory states that an Euler path is p ossible if and only if there are either zero o r tw o no des of o dd degree. Lo oking at Figure 2b, w e see that there are four no des of o dd degree, hence an Euler pat h is not p ossible. The b o a rd in Figure 2 d o es not supp ort a sup er-sw eep. Using the Eule r path theorem we can eliminate ma ny b o ards from having sup er-sw eeps. Hexagonal o r star-shap ed b oards hav e hexagonal symmetry , but do not supp ort sup er-sw eeps 2 Amazingly , the 16-s weep of Figure 1a ca n also b e reached in a peg so litaire problem on a cros s-shap ed bo ard one size larger, the 45- ho le “Wiegleb’s bo ard” [3 ]. The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 4 b ecause the asso ciated sw eep graphs hav e six no des of o dd degree. If w e restrict ourselv es to b oard shap es that are c onvex p olygons (no 240 ◦ or 3 00 ◦ corners) t hing s are particularly simple. F or a conv ex b oard on a triangular lattice, only three board shap es can ha v e a sup er-sw eep: 1. T riangles , and only equilateral t riangles are p ossible on a triangular lattice. Because ev ery no de in the sup er-sw eep g raph has ev en degree, the sup er-sw eep alw a ys b egins and ends at the same b o ard lo cation. 2. Parallelograms , whic h ha ve alternating 60 ◦ and 120 ◦ corners as y ou go ar ound the circumference. The sup er-sw eep mu st b egin at one 120 ◦ corner and end at the other. 3. T rap ezoids , which ha v e t wo 60 ◦ corners follow ed b y t w o 120 ◦ corners. The sup er- sw eep m ust b egin at one 120 ◦ corner and end at the other. These b oards can also b e considered as equilateral triangles with one corner cut off. A rhom bus is a sp ecial case of a parallelogram where all four sides hav e the same length n , w e’ll call this b oa rd Rhombus ( n ). By rotating them 60 ◦ , they become diamond-shap ed and could also b e called Dia mond b oards. In t he remainder of this pap er, we’ll g o into some of t he remark able prop erties of these b oa rds. Figure 3 sho ws the first four r ho m bus b oard sup er-sw eeps. Figure 3: Sup er-sw eeps o n Rhombus ( n ) where n = 3, 5, 7 and 9. These sw eeps hav e lengths of 5, 16, 33, and 56 respective ly , a nd end at the lo w er righ t corner. The length of this sw eep, for o dd n , is (3 n + 1)( n − 1) / 4. Since the total n umber of holes in the b oard is n 2 , this sw eep r emov es nearly 3 / 4 of the p egs on the b o ard. F or n = 3, 5, 7, 9, 11, 1 3, . . . , the sequence of sw eep lengths is 5, 16, 33, 5 6 , 85 , 120, . . . , called the “ rhom bic matc hstic k sequenc e” [4] b ecause it is the n um b er o f matc hstick s needed to cons truct a rhom bus (with ( n − 1) / 2 matchs ticks on a side). 3 Rhombus (6) This 36- ho le b oard has sev eral unu sual prop erties. It is also of a r easonable size for playing b y hand, a nd for computationa l searc hes. This b oard is equiv alen t to t he 6x6 square b oard on a square lattice, with the addition of moves along one diagonal . It is therefore p ossible to pla y this b oard using a c hess or go b oard, althoug h this is not recommended b ecause the symmetry of the b oard is o bscured. F or pla ying b y hand I recommend using par t of a The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 5 Chinese Chec k ers b oard. The ideal b oard for pla ying b y hand is a computer [5], b ecause w e can easily t a k e back mo v es and record mov e sequences. The b oard Rhombus (6) is a n ull-class b oard. F or a definition of this term, see [2] or [5], the imp ortant concept is that only on n ull-class b oards can a complemen t problem b e solv able. Rhombus (6) is the smallest rhom bus b oard on which a complemen t pro blem is solv able, and in fact al l complemen t problems are easily solv able. a1 a2 a3 a4 a5 a6 b1 b2 b3 b4 b5 b6 c1 c2 c3 c4 c5 c6 d1 d2 d3 d4 d5 d6 e1 e2 e3 e4 e5 e6 f1 f2 f3 f4 f5 f6 Figure 4: Rhombus (6) hole co ordinates. P oten tial finishing lo cations for a problem including a maximal sw eep (16-sw eep) are shaded blue. The longest sw eep geometrically p ossible on Rhombus (6) ha s length 16 (as in Figure 3). Can 16-sw eeps o ccur in solutions to p eg solitaire pro blem on this b o ard? Here w e aren’t limiting the 16 -sw eep to b e the last mo v e, but leav e op en the p ossibilit y that it could happ en at any mo v e. W e note tha t there are only a few places where the 16-sw eep can b egin a nd end. It can go f r o m a1 t o e5, b2 to f6, b1 to f5, or any symmetric v aria n t of these. The 16-sw eep can b e the final mov e, or it can b e the second to t he last mo v e, for example the 16 -sw eep can go from a1 to e5, follo wed by f6-d4, or e6-e4 3 . The 16-sw eep can ev en b e the 3rd to the last mov e, from a1 to e5, follo we d by e6- e4 and f5-d3. In Figure 4 , all p oten tial finishing lo cations f or solutions containing a 16 -sw eep are shaded blue. Not all are feasible, the finishing 16-sw eep from b1 to f5 is in fact imp ossible to reac h from any starting v acancy , as discov ered b y computational searc h. How ev er all other configurations of the 16-sw eep can b e reach ed. In fa ct, starting f rom an y v acancy on the b oard, there is a s o lutio n with a maximal sw eep (1 6-sw eep) that finishes with one p eg. This b oard is the o nly one w e kno w of, on a square or triang ula r lattice, with this amazing prop erty . Another un usual prop ert y of Rhombus (6) is that solutions to complemen t pro blems can include ma ximal sw eeps. Here are four different complemen t problems that can b e solve d using a ma ximal sw eep: 1. e5 complemen t : solve with the last mo v e a 16-sw eep. 2. d4 complemen t: solv e with the second to last mov e a 16-sw eep. 3. e4 complemen t : solve with the second to last mov e a 16- sw eep. 4. d3 complemen t: solv e with the third to last mov e a 16-sw eep. 3 Notation: mov es are denoted by the starting and ending co or dinates of the jumps, sepa rated b y da shes. The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 6 These problems are most easily solv ed b y attempting to play bac kw ard, o r equiv alen tly by pla ying forw ard from the complemen t of the b oard p osition b efore the 16-sw eep. All four problems make go o d c hallenges t o solv e b y hand; they are easy to solv e using a computer (pro vided y ou don’t try to solv e them pla ying f orw ard). In Figure 5 w e sho w a solution to problem #4. This solution is in teresting in that after t he third mo ve, the b oard p osition is symmetric ab out the red dashed line. After that mo ves are done in pairs, or are themselv es symmetric, preserving the symmetry up un til t he last tw o mo v es. Figure 5: A symmetric solution to the d3 complemen t (problem #4). This solution has 17 mo v es. Note that more than o ne mo ve is sometimes sho wn b etw een b oard snapshots. An y p eg solitaire problem on this b oard b egins with 35 p egs and finishes with one, so a solution consists of exactly 34 jumps. The n umber of mo ves , ho w ev er, can b e less than this, and an intere sting problem is to find solutions in as few mov es as p ossible. This is differen t from finding solutions with maximal sw eeps, and answ ers are more difficult to o btain. The minimal solution length can b e found using computat io nal searc h metho ds [7 , 8]. If w e tak e into accoun t all p ossible starting and finishing locat io ns for a p eg solitaire problem on this b oard, we find t here are 1 20 distinct pro blems. I hav e only solv ed t he complemen t problems, for there are only 12 of them. Of these 12, I hav e found that 7 can b e solved in a minimum of 1 3 mov es, with the rest requiring 14 mo ves ( see F igure 7 fo r all r esults). A sample 13- mo v e solution is show n in F igure 6. Figure 6: A 13-mov e solution to the c3-complemen t . The last four mo ve s originate from corners, a n un usual prop ert y for a 13-mo v e solution. Note that more than one mo ve is sometimes show n b et we en b oard snapshots. Using the “Merson region” analysis of GPJ #36 [6], it is p ossible to prov e that the 13- mo v e solution in Fig ure 6 is the shortest p ossible. In general, how eve r, we rely on computational The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 7 searc h to establish the minim um. F or m o r e information o n minimal length solutions on Rhombus (6) and ot her b oards see m y web site [5]. The smallest integer (great er than one) that is a triangular num b er and a p erfect square is 36. This is reflected in the fact tha t T riangle (8) and Rhombus (6) b o t h hav e 36 holes. Because of this, it is in teresting to compare the pro p erties of these tw o b o ards. Figure 7 sho ws the minim um length solution of a complemen t pr o blem b y color for each of these b oards. Note that Rhombus (6) in general supports slightly shorter solutions, with none requiring 15 mo ves . 13 moves (magenta) 14 moves (cyan) 15 moves (yellow) Figure 7: Colors indicate the length of the shortest solution to a complemen t problem on T riangle (8) and Rhombus (6) (magen ta = 1 3 mov es, cy an = 1 4 mov es, y ellow = 15 mov es). The Rhombus (6) b oard ha s b een rotated to its “diamond” configuration to sho w the symmetry . 4 Maximal sw eep s on Rhombus (6 i ) In GPJ #3 6 [6], we f ound that maximal sw eeps on T riangle (6) and T riangle (8) could o ccur in p eg solitaire games on these b oa r ds. Although a pro of ha s not b een found, computat io nal results suggest that maximal sw eeps cannot b e reac hed on an y larger triang ular b oards. W e ha ve just sho wn that a maximal sw eep can b e reac hed b y a p eg solitaire problem on Rhombus (6), but what ab out larger rhom bus b oards? One might susp ect that maximal sw eeps w o uld ev en tually b ecome unreachable, as with the triangular b oards. Somewhat remark ably , how eve r, this is no t the case. Theorem F or a n y i > 0, there exists a solution to a p eg solitaire problem on Rhombus (6 i ) where the last mo v e is a maximal sw eep of length (9 i − 1)(3 i − 1). Pr o of: By the F orw ard/Backw ard Theorem [6], it suffices to sho w that the complemen t of some maximal sw eep pattern can b e r educed to one p eg. The swe ep pattern we c ho ose b egins at the upp er left corner, and ends o ne hole up and left from the low er rig ht corner. When w e tak e the complemen t, this results in the b oard p osition of F igure 8. The case i = 1, o r Rhombus (6), has already been solv ed (problem #1 in the previous section). W e will use induction to prov e the general case, starting with i = 2. The solution pro ceeds The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 8 Figure 8: The complemen t of the sw eep pattern on Rhombus (6 i ). Note that the b oar d p osition is symmetric ab out the red da shed line. through t hr ee phases ( A , B and C ). W e apply the mov es o f phase A once, then B ( i − 2) times, follow ed b y phase C once. Figure 9: The eight mo ve s of phase A . Phase A , sho wn in Figure 9, consists of eigh t mov es that clear out the leftmost 6 columns of the b oard and t he upp er 4 rows , except for the la st (rightmost) column. If the b oard is larger than the Rhombus (12) sho wn in Figure 9 , the long m ulti-jump mo v es m ust b e extended accordingly . W e are left with a v ery similar b oard pa ttern a s the one w e started with, just reduced in size. Note, how ev er, that the final b oard p o sition is no longer symmetric. Phase B, sho wn in Figure 10, is nine mo v es that reduce the sw eep pattern by 6 ro ws and 6 columns. As b efore if the b oard is larger than sho wn the multi-jump mo v es are extended. After a pplying phase B j times the leftmost 6 j + 6 columns will b e empt y , and the topmost 6 j + 4 rows will also b e empt y , except for a trail of p egs in the righ tmost column that will b e taken by the final mo v e. Finally , phase C is executed to take the b oar d dow n to a one p eg in the upp er right corner. The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 9 Figure 10: The nine mo v es o f phase B . The mo v es here are sho wn on Rho mbus (15) to sa v e space. Phase B will actually b e applied to b oards at least as large as Rhombus (18). Figure 11: The nine mo v es of phase C ( a lso p ossible in eigh t mov es). The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 10 The nine mo v es of this phase are sho wn in Figure 11. Only in phase C is the fact that the side is divisible by 6 needed. F or only on suc h b oards can the final p eg finish in the upp er righ t hand corner. Putting together all three phases, it o nly tak es 9 i − 1 mo v es to clear the sw eep pattern of Figure 8. T o find the solution ending in the maximal sw eep, w e b egin fr om a v a cancy at the upp er rig ht corner, and execute the jumps of phases A , B and C in exactly the rev erse order. Th us w e execute the jumps in phase C rev ersed, follow ed b y ( i − 2) phase B ’s, and then phase A , all in rev erse order. The long sw eeps become individual jumps, and the solution ending in the maximal sw eep has significan tly more than 9 i − 1 mo ve s. Figure 12: Putting phases A and C together for a bac kw ar d solution on Rhombus (12) . In Figure 12 on Rhombus (12), w e show the complete solution that reduces the complemen t of the sw eep pattern to one p eg. In this case w e need only execute phases A and C , and in Figure 12 the tw o phases ha v e b een in terlea ve d and are no longer visually separate. This reduces the num b er of diagrams to sho w the solution, but the inductiv e nature of the solution b ecomes muc h harder to see. It is unfortunate that a Chinese Chec ke rs b oard is to o small to play this solution on. If y ou can find a large enough b oard, it is in teresting to pla y the mo v es in this solution in exactly the rev erse order, and w atc h as the sw eep po sition magically app ears. The final sw eep in the rev ersed solution has length 85. An integer that is a p erfect square and a triangular num b er is called a square triangular n um b er [4]. As w as the case with Rhombus (6) and T riang le (8), which b o th ha ve 36 holes, eac h square t riangular n umber corresp onds to a rhom bus and triangular b oard of the same size. If the side of the rhombus b oard is divisible b y 6, and the side of the tr ia ngular b oar d b y 12, then b o t h ha v e long sw eep finishes b y the ab o ve a nalysis and GPJ #36 [6]. After t he 36-hole b oards, the next time this o ccurs is with Rhombus (204) and T riangle (288), whic h b ot h hav e 41 , 616 holes. By our inductiv e argumen ts, we can construct solutio ns to p eg solitaire problems on these b oards that finish with sw eeps of length 30 , 8 05 and 30 , 79 3, The Games and Puzzles Journal—Issue 41, Septem b er-Octob er 20 05 11 resp ectiv ely . The next larger suc h b oards are R hombus (235 , 416) and T riangle (33 2 , 928), b oards with o v er 55 billion holes, that can finish with sw eeps ov er 41 billion in length. 5 Conclus ions P eg solitaire on a triangular lattice is a fascinating game, and o ne that has not b een studied to the extent that “normal” (square lattice) peg solitaire has. This pap er, and GPJ #36 [6] ha v e sho wn t hat triangula r la ttice p eg solitaire is w ell suited for inductive argumen ts. Inductiv e argumen ts ha v e also b een used to create an algorithm for triangular b oards of an y size that can reduce a n y ( solv able) single v acancy do wn to one p eg [8]. I b eliev e inductiv e argumen ts are p o ssible for square lattice b oa r ds, but the reduction from 6 j ump directions to 4 seems t o mak e suc h a rgumen ts more difficult. While the results in Section 3 for Rhombus (6) we re obtained by computational searc h, the long sw eep finishes on Rhombus (6 i ) w ere found b y hand. The computer w a s still of significan t help, but only in prov iding an interface to pla y the game on the large b oards required. There remain man y una nswe red questions regarding rhombus and tr ia ngular b oards. Can maximal swe eps b e r eached on p eg solitaire problems on Rhombus (2 i )? I ha ve b een able to answ er this question in the a ffirmativ e for 2 ≤ i ≤ 9 . It w ould also b e nice to prov e that maximal sw eeps are not reac hable in p eg solitaire problems on triangular b oa rds larger than T riangle (8) (or find a coun terexample). References [1] E . Berlek a mp, J . Co nw ay , and R. Guy , Winning Ways for your Mathematic al Plays , A K Peters, V ol. 2: 695–7 34 (1982 edition) or V ol. 4: 8 03–84 1 (2004 edition). [2] J . Bea sley , The Ins and Outs of Pe g S olitair e , Oxfor d Univ. Press, Oxford, New Y ork, 1992. [3] G. B ell and J. Beas ley , New pr o blems on old solitaire b oar ds, Boar d Ga me Studies, 8 (20 07) (to app ear), http://w ww.bo ardgamesstudies.org , arXiv:ma th/06 11091 [ math. CO] [4] N. Sloa ne , The On-Line E nc y clop edia o f In teger Sequences, rho mb ic matchstic k num be r s a re se - quence A04 5944 , square triangular num b ers are sequenc e A00111 0 http:/ /www. research.att.com/ %7 enjas/ seque nces/ [5] h ttp:/ /www. geocities.com/gibell.geo/pegsolitaire/ (should this web a ddress c hang e in the future I sugg est a search on keywords: “ triangular peg solitaire George Bell” ) [6] G. Bell, T riangula r Peg Solitaire Unlimited, The Games and P uzzles Journal, 36 , Dec. 2 0 04 http:/ /www. gpj.connectfree.co.uk/gpjr.htm , arXi v:071 1.0486 [math.C O] [7] G. B e ll, Diagonal Peg Solitair e, INTEGERS Ele ctr onic Journal of Combinatorial Numb er The ory , V ol 7, G1, 2007 , arXi v:mat h/060 6122 [mat h.CO] [8] G. B ell, Solving T riangula r Peg Solitaire, submitted to The Ele ctr onic Journal of Combinatorics , arXiv: math/ 0703865 [ math. CO]